560 nha

560

nha ban

chuc ban hoc gioi

20 x 1 + 20 x 2 + 20 x 3 + 20 x4 + 20 x5 + 20 x 6 + 20 x7

 = 20 + 40 + 60 + 80 + 100 + 120 + 140 

= 560 

k nha bn

=20x(1+2+3+4+5+6+7+8+9+10)

=20x55=1100

=1100

k nha

Bạn viết đề thiếu trầm trọng quá !!!

Đáp án: thiếu đề

@#@

mời bn xem xét lại đề bài.

~hok tốt~

20x1+20x2+20x3+20x4+20x5+20x6= 20x(1+2+3+4+5+6)= 20 x 21 = 420

Còn câu 2 mik thấy khó quá cho mik xin lỗi nha

=20+40+60+80+100+120

=420

Con thứ nhất: = 20.(1+2+3+4+5+6)

                      =420

Con 2 để mình nghĩ nhé!

\(20+\frac{20}{10}+87+88+90+\frac{10\times100}{10}\times\frac{800}{1000}\)

\(=20+2+175+90+100+0,8\)

\(=22+265+100,8\)

\(=287+100,8\)

\(=387,8\)

a, Ta có: \(\frac{2001}{2002}=\frac{2002-1}{2002}=\frac{2002}{2002}-\frac{1}{2002}=1-\frac{1}{2002}\)

\(\frac{2000}{2001}=\frac{2001-1}{2001}=\frac{2001}{2001}-\frac{1}{2001}=1-\frac{1}{2001}\)

Vì \(\frac{1}{2002}< \frac{1}{2001}\Rightarrow1-\frac{1}{2002}>1-\frac{1}{2001}\Rightarrow\frac{2001}{2002}>\frac{2000}{2001}\)

b, Ta có: \(\left(\frac{1}{80}\right)^7>\left(\frac{1}{81}\right)^7=\left(\frac{1}{3^4}\right)^7=\left(\frac{1}{3}\right)^{28}=\frac{1}{3^{28}}\)

\(\left(\frac{1}{243}\right)^6=\left(\frac{1}{3^5}\right)^6=\left(\frac{1}{3^5}\right)^6=\frac{1}{3^{30}}\)

Vì \(\frac{1}{3^{28}}>\frac{1}{3^{30}}\Rightarrow\left(\frac{1}{81}\right)^7>\left(\frac{1}{243}\right)^6\Rightarrow\left(\frac{1}{80}\right)^7>\left(\frac{1}{243}\right)^6\)

c, Ta có: \(\left(\frac{3}{8}\right)^5=\frac{3^5}{\left(2^3\right)^5}=\frac{243}{2^{15}}>\frac{243}{3^{15}}>\frac{125}{3^{15}}=\frac{5^3}{\left(3^5\right)^3}=\frac{5^3}{243^3}=\left(\frac{5}{243}\right)^3\)

Vậy \(\left(\frac{3}{8}\right)^5>\left(\frac{5}{243}\right)^3\)

d, Ta có: \(\frac{2011}{2012}>\frac{2011}{2012+2013}\)

\(\frac{2012}{2013}>\frac{2012}{2012+2013}\)

\(\Rightarrow\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{2012+2013}+\frac{2012}{2012+2013}=\frac{2011+2012}{2012+2013}\)

e, \(C=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=\frac{20^{10}-1}{20^{10}-1}+\frac{2}{2^{10}-1}=1+\frac{2}{2^{10}-1}\)

\(D=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=\frac{20^{10}-3}{20^{10}-3}+\frac{2}{2^{10}-3}=1+\frac{2}{2^{10}-3}\)

Vì \(\frac{2}{10^{10}-1}< \frac{2}{10^{10}-3}\Rightarrow1+\frac{2}{10^{10}-1}< 1+\frac{2}{10^{10}-3}\Rightarrow C< D\)

g, \(G=\frac{10^{100}+2}{10^{100}-1}=\frac{10^{100}-1+3}{10^{100}-1}=\frac{10^{100}-1}{10^{100}-1}+\frac{3}{10^{100}-1}=1+\frac{3}{10^{100}-1}\)

\(H=\frac{10^8}{10^8-3}=\frac{10^8-3+3}{10^8-3}=\frac{10^8-3}{10^8-3}+\frac{3}{10^8-3}=1+\frac{3}{10^8-3}\)

Vì \(\frac{3}{10^{100}-1}< \frac{3}{10^8-3}\Rightarrow1+\frac{3}{10^{100}-1}< 1+\frac{3}{10^8-3}\Rightarrow G< H\)

h, Vì E < 1 nên:

\(E=\frac{98^{99}+1}{98^{89}+1}< \frac{98^{99}+1+97}{98^{89}+1+97}=\frac{98^{99}+98}{98^{89}+98}=\frac{98\left(98^{98}+1\right)}{98\left(98^{88}+1\right)}=\frac{98^{98}+1}{98^{88}+1}=F\)

Vậy E = F

\(\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+...+\dfrac{1}{\left(x+2\right)\left(x+5\right)}=\dfrac{3}{20}\)

\(\Rightarrow\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{\left(x+2\right)\left(x+5\right)}=\dfrac{3}{20}\)

\(\Rightarrow\dfrac{1}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{\left(x+2\right)\left(x+5\right)}\right)=\dfrac{3}{20}\)

\(\Rightarrow\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+..+\dfrac{3}{\left(x+2\right)\left(x+5\right)}=\dfrac{9}{20}\)

\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{x+2}-\dfrac{1}{x+5}=\dfrac{9}{20}\)

\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+5}=\dfrac{9}{20}\)

\(\Rightarrow\dfrac{1}{x+5}=\dfrac{1}{2}-\dfrac{9}{20}\)

\(\Rightarrow\dfrac{1}{x+5}=\dfrac{1}{20}\)

\(\Rightarrow x+5=20\)

\(\Rightarrow x=20-5\)

\(\Rightarrow x=15\)

8:

\(A=\dfrac{20^{10}-1+2}{20^{10}-1}=1+\dfrac{2}{20^{10}-1}\)

\(B=\dfrac{20^{10}-3+2}{20^{10}-3}=1+\dfrac{2}{20^{10}-3}\)

mà 20^10-1>20^10-3

nên A<B

làm lại

ta có : \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)

=>\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)

=>\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+....+\frac{3}{\left(x+2\right)\left(x+5\right)}=\frac{9}{20}\)

=>\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{x+2}-\frac{1}{x+5}=\frac{9}{20}\)

=>\(\frac{1}{2}-\frac{1}{x+5}=\frac{9}{20}\)

=>\(\frac{1}{x+5}=\frac{1}{2}-\frac{9}{20}\)

=>\(\frac{1}{x+5}=\frac{1}{20}\)

=>\(x+5=20\)

=>\(x=20-5\)

=>\(x=15\)

ta có : \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+....+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)

=>\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{3}{20}\)

=>\(3.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+....+\frac{1}{\left(x+2\right)\left(x+3\right)}\right)=3.\frac{3}{20}\)

=>\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+....+\frac{3}{\left(x+2\right)\left(x+3\right)}=\frac{9}{20}\)

=>\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{x+2}-\frac{1}{x+3}=\frac{9}{20}\)

=>\(\frac{1}{2}-\frac{1}{x+3}=\frac{9}{20}\)

=>\(\frac{1}{x+3}=\frac{1}{2}-\frac{9}{20}\)

=>\(\frac{1}{x+3}=\frac{1}{20}\)

=>\(x+3=20\)

=>\(x=20-3\)

=>\(x=17\)

\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\\ \frac{1}{3}\left(\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+...+\frac{3}{\left(x+2\right)\left(x+5\right)}\right)=\frac{3}{20}\\ \frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{\left(x+2\right)\left(x+5\right)}\right)=\frac{3}{20}\\ \frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x+2}-\frac{1}{x+5}\right)=\frac{3}{20}\\ \frac{1}{3}\left(\frac{1}{2}-\frac{1}{x+5}\right)=\frac{3}{20}\\ \frac{1}{2}-\frac{1}{x+5}=\frac{3}{20}:\frac{1}{3}\\ \frac{1}{2}-\frac{1}{x+5}=\frac{3}{20}\cdot3\\ \frac{1}{2}-\frac{1}{x+5}=\frac{9}{20}\\ \frac{1}{x+5}=\frac{9}{20}+\frac{1}{2}\\ \frac{1}{x+5}=\frac{9}{20}+\frac{10}{20}\\ \frac{1}{x+5}=\frac{19}{20}\\ \frac{19}{19\left(x+5\right)}=\frac{19}{20}\\ \Rightarrow19\left(x+5\right)=20\\ 19x+95=20\\ 19x=20-95\\ 19x=-75\\ x=\left(-75\right):19\\ x=\frac{75}{19}\)

Vậy \(x=\frac{75}{19}\)